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Find the domain of the function given: $f(n)=\frac{\sqrt{n-2}}{n^2+25}$



The domain is:

A$n \leq 2$
B$n

eq 2$

C$n \leq-1$
D$n \geq 2$
EAll real numbers

Verified Answer :

  • The radicand of the square root must be non-negative, requiring \(n - 2 \geq 0\), which simplifies to \(n \geq 2\).
  • The denominator \(n^2 + 25\) is strictly positive for all real numbers \(n\), so it imposes no additional restrictions.
  • Combining these conditions, the domain of the function is \(n \geq 2\).
  • The correct option is D: \(\boxed{n \geq 2}\).

Explanation

  1. Analysis of the Function and its Components

To find the domain of the function \(f(n) = \frac{\sqrt{n-2}}{n^2 + 25}\), we must identify all real values of \(n\) for which the expression is defined. A function is defined if all its constituent operations are valid within the real number system. In this case, we have two primary conditions to consider: the square root and the fraction.

  1. Addressing the Square Root Condition

First, consider the square root term, \(\sqrt{n-2}\). For the square root of a real number to be defined, the radicand (the expression inside the root) must be greater than or equal to zero. Therefore, we set up the inequality:
\[n - 2 \geq 0\]
Adding 2 to both sides gives:
\[n \geq 2\]
This means the function is only defined for real numbers \(n\) that are greater than or equal to 2.

  1. Addressing the Denominator Condition

Next, consider the denominator of the fraction, \(n^2 + 25\). A fraction is undefined when its denominator is zero. We must ensure that:
\[n^2 + 25 eq 0\]
Subtracting 25 from both sides gives:
\[n^2 eq -25\]
Since the square of any real number \(n\) is always non-negative (\(n^2 \geq 0\)), it follows that \(n^2 + 25 \geq 25\) for all real numbers \(n\). Thus, the denominator \(n^2 + 25\) is always positive and can never be zero. Therefore, there are no additional restrictions on \(n\) imposed by the denominator.

  1. Final Conclusion

Combining the conditions from the previous steps, the only restriction on \(n\) is that \(n \geq 2\). This covers all values for which the square root is defined, and since the denominator is never zero, the entire function \(f(n)\) is well-defined for all \(n\) in this interval. Thus, the domain of the function is \(n \geq 2\). Comparing this result with the given options, it corresponds to option D.

Examples

In physics and engineering, domains are essential when modeling physical processes. For instance, if a variable \(n\) represents the time elapsed since an object was launched, time cannot be negative (\(n \geq 0\)). Similarly, if the speed of a particle follows the function \(v(n) = \frac{\sqrt{n-2}}{n^2 + 25}\), the particle's movement only begins at time \(n=2\). Understanding the domain ensures that we only calculate values where the physical system is valid, preventing nonsensical results like imaginary velocities or division by zero.

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